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\title{\heiti\zihao{2} 习题18.2}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{计算下列第二型曲面积分}
\subsection{$\iint\limits_{\Sigma}\left(x^{2}-y\right) \mathrm{d} y \mathrm{~d} z-y \mathrm{~d} z \mathrm{~d} x+\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y$, 其中 $\Sigma$ 为旋转拋物面 $z=x^{2}+y^{2}, z \leqslant 1$,方向取外侧}
\textbf{解}\quad
法向量为$\vec{n}=(-z_x,-z_y,1)=(-2x,-2y,1)$,外法向量为$-\vec{n}$.
$$
	\begin{aligned}
		I & =\iint\limits_{\Sigma}\left(x^{2}-y\right) \mathrm{d} y \mathrm{~d} z-y \mathrm{~d} z \mathrm{~d} x+\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \\
		  & =\iint\limits_{\Sigma}2x^3-2xy-3y^2-x^2 \mathrm{d} x \mathrm{~d} y                                                                                      \\
		  & =\int_0^{2\pi}\mathrm{d}\theta\int_0^1(2\rho^3\cos^3\theta-2\rho^2\cos\theta\sin\theta-3\rho^2\sin^2\theta-\rho^2\cos^2\theta)\rho\mathrm{d}\rho              \\
		  & =\pi
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma} z \mathrm{~d} y \mathrm{~d} z+x \mathrm{~d} z \mathrm{~d} x+y \mathrm{~d} x \mathrm{~d} y$, 其中 $\Sigma$ 为柱面 $x^{2}+y^{2}=1$ 被 $z=0, z=h$ 所截部分, 方向取外侧}
\textbf{解}\quad
柱坐标变换.
$$
	\begin{aligned}
		\iint\limits_{\Sigma} z \mathrm{~d} y \mathrm{~d} z+x \mathrm{~d} z \mathrm{~d} x+y \mathrm{~d} x \mathrm{~d} y & =\iint\limits_Dz\cos\theta+\cos\theta\sin\theta\mathrm{d}\varphi\mathrm{d}z       \\
		                                                                                                                & =\int_0^{2\pi}\mathrm{d}\theta\int_0^hz\cos\theta+\cos\theta\sin\theta\mathrm{d}z \\
		                                                                                                                & =0
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma}\left(z^{2}+x\right) \mathrm{d} y \mathrm{~d} z-z \mathrm{~d} x \mathrm{~d} y$, 其中 $\Sigma$ 为 $z=\dfrac{1}{2}\left(x^{2}+y^{2}\right)$ 介于平面 $z=0, z=2$ 之间的部分的下侧}
\textbf{解}\quad
外法向量$\vec{n}=(x,y,-1)$.$P=z^2+x,Q=0,R=-z$.
$$
	\begin{aligned}
		\iint\limits_{\Sigma}\left(z^{2}+x\right) \mathrm{d} y \mathrm{~d} z-z \mathrm{~d} x \mathrm{~d} y & =\iint\limits_{\Sigma}xz^2+x^2+z\mathrm{d}x\mathrm{d}y                                                                     \\
		                                                                                                   & =\iint\limits_{\Sigma}x\dfrac{(x^2+y^2)^2}{4}+x^2+\dfrac{(x^2+y^2)}{2}\mathrm{d}x\mathrm{d}y                               \\
		                                                                                                   & =\int_0^{2\pi}\mathrm{d}\theta\int_0^2\rho^2\cos\theta\dfrac{\rho^4}{4}+\rho^3\cos^2\theta+\dfrac{\rho^3}{2}\mathrm{d}\rho \\
		                                                                                                   & =8\pi
	\end{aligned}
$$
\subsection{$\iint\limits_{\Sigma} z \mathrm{~d} x \mathrm{~d} y$, 其中 $\Sigma$ 为球面 $x^{2}+y^{2}+z^{2}=a^{2}$ 在第一卦限的部分与各坐标面所围成立体表面的外侧}
\textbf{解}$1^{\circ}$\quad
$$
	\begin{aligned}
		\iint\limits_{\Sigma} z \mathrm{~d} x \mathrm{~d} y & =\iint\limits_\Sigma\sqrt{a^2-x^2-y^2}\mathrm{d}x\mathrm{d}y              \\
		                                                    & =\int_0^{2\pi}\mathrm{d}\theta\int_0^a\sqrt{a^2-\rho^2}\rho\mathrm{d}\rho \\
		                                                    & =\dfrac{a^2\pi}{6}
	\end{aligned}
$$

\textbf{解}$2^{\circ}$\quad
极坐标变换:$\left\{\begin{array}{l}
		x=a\sin\varphi\cos\theta \\
		y=a\sin\varphi\sin\theta \\
		z=a\cos\varphi
	\end{array}\right.$.则外法向量:$\vec{n}=(\sin^2\varphi\cos\theta,\sin^2\varphi\sin\theta,\sin\varphi\cos\varphi)$
$$
	\begin{aligned}
		\iint\limits_{\Sigma} z \mathrm{~d} x \mathrm{~d} y & =a^2\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\pi/2}\sin\varphi\cos^2\varphi\mathrm{d}\varphi \\
		                                                    & =\dfrac{a^2\pi}{6}
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma}\left(x^{2}+y^{2}\right) \mathrm{d} y \mathrm{~d} z+z \mathrm{~d} x \mathrm{~d} y$, 其中 $\Sigma$ 为柱面 $x^{2}+y^{2}=R^{2}$ 与 $z=0, z=H(H>0)$ 所围柱体表面的外侧}
\textbf{解}\quad
柱坐标变换:外法向量$\vec{n}=(R\cos\theta,R\sin\theta,0)$.
$$
	\begin{aligned}
		\iint\limits_{\Sigma}\left(x^{2}+y^{2}\right) \mathrm{d} y \mathrm{~d} z+z \mathrm{~d} x \mathrm{~d} y & =\iint\limits_{\Sigma}R^3\cos\theta\mathrm{d}\theta\mathrm{d}z \\
		                                                                                                       & =\int_0^{2\pi}\mathrm{d}\theta\int_0^HR^3\cos\theta\mathrm{d}z \\
		                                                                                                       & =0
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma}(y-z) \mathrm{d} y \mathrm{~d} z+(z-x) \mathrm{d} z \mathrm{~d} x+(x-y) \mathrm{d} x \mathrm{~d} y$, 其中 $\Sigma$ 为锥面 $z^{2}=x^{2}+y^{2}(0 \leqslant z \leqslant b)$ 的外侧.}
\textbf{解}\quad
外法向量$\vec{n}=(\dfrac{x}{\sqrt{x^2+y^2}},\dfrac{y}{\sqrt{x^2+y^2}},-1)$.
$$
	\begin{aligned}
		I & =\iint\limits_{\Sigma}(y-z) \mathrm{d} y \mathrm{~d} z+(z-x) \mathrm{d} z \mathrm{~d} x+(x-y) \mathrm{d} x \mathrm{d} y                     \\
		  & =\iint\limits_{\Sigma}\dfrac{x(y-\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}+\dfrac{y(\sqrt{x^2+y^2}-x)}{\sqrt{x^2+y^2}}+y-x \mathrm{d} x \mathrm{d} y \\
		  & =2\iint\limits_{\Sigma}y-x\mathrm{d} x \mathrm{d} y                                                                                         \\
		  & =0(\text{对称性,对}x,y\text{积分的值应相等,都为}0)
	\end{aligned}
$$

\section{设磁场强度为 $E(x, y, z)=(x^2,y^2,z^2)$, 求从球内出发通过上球面 $x^{2}+y^{2}+z^{2}=a^{2}, z \geqslant 0$ 的磁通量.}
\textbf{解}\quad
记$\boldsymbol{E}(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))$.
上球面:
$$
	\begin{aligned}
		z=\sqrt{a^2-x^2-y^2}
	\end{aligned}
$$
外法向量$\vec{n}=\left(\dfrac{x}{\sqrt{a^2-x^2-y^2}},\dfrac{y}{\sqrt{a^2-x^2-y^2}},1\right)$.所以
$$
\begin{aligned}
	\phi&=\iint\limits_D\dfrac{xP}{\sqrt{a^2-x^2-y^2}}+\dfrac{yQ}{\sqrt{a^2-x^2-y^2}}+R\mathrm{d}x\mathrm{d}y\\
	&=\iint\limits_D\left(\dfrac{x^3+y^3}{\sqrt{a^2-x^2-y^2}}+a^2-x^2-y^2\right)\mathrm{d}x\mathrm{d}y\\
	&=\dfrac{a^4\pi}{2}
\end{aligned}
$$









\end{document}